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The cosmic rays of highest energy are protons that have kinetic energy on the order of 10¹³ MeV (a) As measured in the proton's frame, what time interval would a proton of this energy require to travel across the Milky Way galaxy, which has a proper diameter ≅10⁵ly ?

Sagot :

The relativistic time it takes the proton to travel the milky way is 9.877 × 10⁻⁷ s

How to find the time it takes the proton to travel the milky way?

Since the question has to do with relativistic motion, we need to find the relativistic time it takes the proton to travel the milky way. First, we use the equation for relativistic kinetic energy, we have

K = mc(γ - 1) where

  • K = kinetic energy of proton = 10¹³ MeV = 10¹³ × 1.602 × 10⁻¹³ J = 1.602 J,
  • m = rest mass of proton = 1.6726 × 10⁻²⁷ kg,
  • c = speed of light = 3 × 10⁸ m/s and
  • γ = 1/[√(1 - (v/c)²] where
  • v = speed of proton

So, making γ subject of the formula, we have

γ  = K/mc + 1

Substituting the values of the variables into the equation, we have

γ  = K/mc + 1

γ  = 1.602J/(1.6726 × 10⁻²⁷ kg × 3 × 10⁸ m/s) + 1

γ  = 1.602J/(5.0178 × 10⁻¹⁹ J) + 1

γ  = 0.3193 × 10¹⁹ + 1

γ  = 3.193 × 10¹⁸

The distance moved by the proton

Now since the proper diameter of the milky way d = 10⁵ ly, since the length moved by the proton is going to be contracted, the actual distance moved by the proton is

d' = d/γ

= 10⁵ ly/3.193 × 10¹⁸

= 0.3132 × 10⁻⁻¹³ ly

= 0.3132 × 10⁻⁻¹³ × 9.46 × 10¹⁵ m

= 2.963 × 10² m

The velocity of the proton

Next we need to find the velocity of the proton from

γ = 1/[√(1 - (v/c)²]

3.193 × 10¹⁸ = 1/[√(1 - (v/c)²]

√(1 - (v/c)² = 1/3.193 × 10¹⁸

√(1 - (v/c)² = 0.3132 × 10¹⁸

1 - (v/c)² = (0.3132 × 10⁻¹⁸)²

1 - (v/c)² = 0.0981 × 10⁻³⁶

(v/c)² = 1 - 0.0981 × 10⁻³⁶

(v/c)² = 1 - 0.0000000000000000000000000000000000981

(v/c)² = 0.9999999999999999999999999999999999019

(v/c)² ≅ 0.99999

(v/c) = √0.99999

v/c = 0.999994999987

v = 0.999994999987c

The time it takes the proton to travel the milky way galaxy

So, the time it takes the proton to travel the milky way is t = d/v

= 2.963 × 10² m/0.999994999987c

= 2.963 × 10² m/2.99998499996 × 10⁸ m/s

= 0.9877 × 10⁻⁶ s

= 9.877 × 10⁻⁷ s

So, the time it takes the proton to travel the milky way is 9.877 × 10⁻⁷ s

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